Approach by Solar Plexsus: Consider the Diophantine system of equations $ (1) \;\; 100a + b = m^2$ $ (2) \;\; 201a + b = n^2$ where $ 10 \leq a,b < 100$ and $ 32 \leq m < n < 100$. By subtracting (1) from (2), we obtain $ (3) \;\; 101a = n^2 - m^2 = (n - m)(n + m)$. Since 101 is a prime and $ 0 < n - m < 101$, $ 101 | n + m$ by (3). This combined with the fact that $ 0 < n + m < 2*101$ implies that $ n + m = 101$, i,e $ n = 101 - m$. According to (3) $ 101a = (101 - m)^2 - m^2 = 101^2 - 2 \cdot 101m$, i.e. $ a = 101 - 2m$. From (2) we obtain $ b = n^2 - 201a = (101 - m)^2 - 201(101 - 2m) = m^2 + 200m - 10100.$ Now $ 10 \leq b < 100$ implies that $ m = 42$, thus $ 100a + b = 42^2 = 1764$. by (1). In other words, the only solution of system of equations (1)-(2) is $ (a,b) = (17,64).$ Approach by jmerry: The only such pair is $ a = 17,b = 64$, with $ 100a + b = 42^2 = 1764$ and $ 201a + b = 59^2 = 3781$ Let $ u^2 = 201a + b$ and $ v^2 = 100a + b$. Then by the four-digit restriction, $ 32\le u < v \le 99$. Now $ 101a = u^2 - v^2 = (u - v)(u + v)$. Since 101 is prime and $ u - v < 101$, $ 101|u + v$. Also, $ u + v < 202$, so $ u + v = 101$, leaving $ u - v = a$. Now we have $ u = \frac {101 + a}{2}$, so $ \left(\frac {101 + a}{2}\right)^2 = 201a + b$ $ 101^2 + 202a + a^2 = 804a + 4b$ $ 301^2 - 101^2 + 4b = a^2 - 602a + 301^2 = \left(301 - a\right)^2$ Therefore $ 200*402 + 4b = 4(100*201 + b)$ is a perfect square. Since squares greater than 10000 are at least 200 apart, there is at most one solution. We have $ \sqrt {100*201 + b} > 100\sqrt {2}$ by a small margin; guessing $ \sqrt {100*201 + b} = 142$ gives $ 142^2 = 20164 = 100*201 + 64$, so $ b = 64$ and $ 301 - a = 2*142 = 284$ so $ a = 17$.

Sorry to revive but the problem is written wrong, it actually puts the restrinction that both $n^2$ and $m^2$ are four digit numbers. Here is the original paper (in Spanish). http://www.oei.es/oim/xixd2.PDF

Above: Yes, the problem does say that $n^2$ and $m^2$ must be four digit numbers. However, this information is unnecessary because, as established by the answers above, the only solution gives $n^2$ and $m^2$ with four digits. Indeed, $a = 17$ and $b = 64$ yields $100a + b = 42^2 = 1764$ and $201a + b = 59^2 = 3481$.

orl wrote: Approach by Solar Plexsus: Consider the Diophantine system of equations $ (1) \;\; 100a + b = m^2$ $ (2) \;\; 201a + b = n^2$ where $ 10 \leq a,b < 100$ and $ 32 \leq m < n < 100$. By subtracting (1) from (2), we obtain $ (3) \;\; 101a = n^2 - m^2 = (n - m)(n + m)$. Since 101 is a prime and $ 0 < n - m < 101$, $ 101 | n + m$ by (3). This combined with the fact that $ 0 < n + m < 2*101$ implies that $ n + m = 101$, i,e $ n = 101 - m$. According to (3) $ 101a = (101 - m)^2 - m^2 = 101^2 - 2 \cdot 101m$, i.e. $ a = 101 - 2m$. From (2) we obtain $ b = n^2 - 201a = (101 - m)^2 - 201(101 - 2m) = m^2 + 200m - 10100.$ Now $ 10 \leq b < 100$ implies that $ m = 42$, thus $ 100a + b = 42^2 = 1764$. by (1). In other words, the only solution of system of equations (1)-(2) is $ (a,b) = (17,64).$ Approach by jmerry: The only such pair is $ a = 17,b = 64$, with $ 100a + b = 42^2 = 1764$ and $ 201a + b = 59^2 = 3781$ Let $ u^2 = 201a + b$ and $ v^2 = 100a + b$. Then by the four-digit restriction, $ 32\le u < v \le 99$. Now $ 101a = u^2 - v^2 = (u - v)(u + v)$. Since 101 is prime and $ u - v < 101$, $ 101|u + v$. Also, $ u + v < 202$, so $ u + v = 101$, leaving $ u - v = a$. Now we have $ u = \frac {101 + a}{2}$, so $ \left(\frac {101 + a}{2}\right)^2 = 201a + b$ $ 101^2 + 202a + a^2 = 804a + 4b$ $ 301^2 - 101^2 + 4b = a^2 - 602a + 301^2 = \left(301 - a\right)^2$ Therefore $ 200*402 + 4b = 4(100*201 + b)$ is a perfect square. Since squares greater than 10000 are at least 200 apart, there is at most one solution. We have $ \sqrt {100*201 + b} > 100\sqrt {2}$ by a small margin; guessing $ \sqrt {100*201 + b} = 142$ gives $ 142^2 = 20164 = 100*201 + 64$, so $ b = 64$ and $ 301 - a = 2*142 = 284$ so $ a = 17$. but if a = b = 99, then n > 100 ...

If a pair $(a,b)$ satisfies we have $100a +b=x^2$, $201a+b = y^2$, where $x,y \in \mathbb{Z}$ such that $0 < x < y < 100$. From which we easily have $101a=(y-x)(y+x)$, which implies $(y-x)=a$ and $(y+x)=101$, from which subtracting the latter from $ y+x=101$ we have $x=\frac{101-a}{2}$ Plugging this into $100a+b=x^2$ gives $a^2 - 602a +(101^2-4b)=0$. Then since a has two digits we have $a=301 \pm 2 \sqrt{20100+b}$. For a to be an integer we have that $20100+b$ is a perfect square, like $10 \leq b \leq 99$, then $20110 \leq 20100+b < 20200$. Since $141^2=19881$ and $143^2=20449$, we necessarily have $20100+b=142^2=142^2=20164$, from which $b=64$, which implies $a= 301-2\sqrt {142^2}=17$. Where you can see that $(a;b)=(17;64)$ satisfies.